Question

Give plausible explanation for each of the following:
(i) Cyclohexanone forms cyanohydrin in good yield but 2,2,6-trimethylcyclohexanone does not.
(ii) There are two $N{H}_2$ groups in semicarbazide. However, only one is involved in the formation of semicarbazones.
(iii) During the preparation of esters from a carboxylic acid and an alcohol in the presence of an acid catalyst, the water or the ester should be removed as soon as it is formed.

Answer 1

(i) In 2,2,6-trimethylcyclohexanone, three methyl groups are present at alpha position with respect to carbonyl group. Hence, the attack of cyanide nucleophile is sterically hindered. In cyclohexanone, there is little steric hinderance. Hence, cyclohexanone forms cyanohydrin in good yield but 2,2,6-trimethylcyclohexanone does not.

(ii) In senicarbazide, out of two $-N{H}_2$ groups, one is involved in resonance with amide carbonyl. The lone pair of electron on N is delocalized through resonance and cannot be donated to a suitable electrophile. Hence, this $-N{H}_2$ group cannot act as a nucleophile. Hence, out of two $-N{H}_2$ groups in semicarbazide, only one is involved in the formation of semicarbazones.

(iii) During the preparation of esters from a carboxylic acid and an alcohol in the presence of an acid catalyst, the water or the ester should be removed as soon as it is formed. This ensures that the equilibrium will shift in the forward direction and more and more of ester will be formed.
Note: The formation of an ester by condensation of carboxylic acid and alcohol is a reversible reaction.